0A: accelerating
AB: constant
BC: decelerating
CD:at rest
DE:accelerating
EF: constant
hope this helps
Answer:
(A) Work done will be 87.992 KJ
(B) Work done will be 167.4 KJ
Explanation:
We have given mass of methane m = 4.5 gram = 0.0045 kg
Volume occupies ![V_1=12.7dm^3=12.7liters](https://tex.z-dn.net/?f=V_1%3D12.7dm%5E3%3D12.7liters)
And volume is increased by
so ![V_2=12.7+3.3=16liters](https://tex.z-dn.net/?f=V_2%3D12.7%2B3.3%3D16liters)
Temperature T = 310 K
Pressure is given as 200 Torr = 26664.5 Pa
(a) At constant pressure work done is given by
![W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj](https://tex.z-dn.net/?f=W%3DP%28V_2-V_1%29%3D26664.5%5Ctimes%20%2816-12.7%29%3D87992.85J%3D87.992kj)
(b) At reversible process work done is given by ![W=nRTln\frac{V_2}{V_1}](https://tex.z-dn.net/?f=W%3DnRTln%5Cfrac%7BV_2%7D%7BV_1%7D)
We have given mass = 4.5 gram
Molar mass of methane = 16
So number of moles ![n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bmass%5C%20in%5C%20gram%7D%7Bmol%3Bar%5C%20mass%7D%3D%5Cfrac%7B4.5%7D%7B16%7D%3D0.28125)
So work done ![W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J](https://tex.z-dn.net/?f=W%3D0.28125%5Ctimes%208.314%5Ctimes%20310ln%5Cfrac%7B16%7D%7B12.7%7D%3D167.4J)
Answer:
electric motors is the answer
Answer:
White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
Complete Question
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:
![\theta=274rev](https://tex.z-dn.net/?f=%5Ctheta%3D274rev)
Explanation:
From the question we are told that:
Angular velocity ![\omega=510rpm](https://tex.z-dn.net/?f=%5Comega%3D510rpm)
Mass ![m=40.kg](https://tex.z-dn.net/?f=m%3D40.kg)
Diameter d ![75=>0.75m](https://tex.z-dn.net/?f=75%3D%3E0.75m)
Off Time ![t=40.0s](https://tex.z-dn.net/?f=t%3D40.0s)
Oscillation at Power off ![N=210](https://tex.z-dn.net/?f=N%3D210)
Generally the equation for Angular displacement is mathematically given by
![\theta_{\infty}=\frac{w+w_0}{t}t](https://tex.z-dn.net/?f=%5Ctheta_%7B%5Cinfty%7D%3D%5Cfrac%7Bw%2Bw_0%7D%7Bt%7Dt)
![w=\frac{2*\theta_{\infty}}{t}-w_0](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B2%2A%5Ctheta_%7B%5Cinfty%7D%7D%7Bt%7D-w_0)
![w=\frac{28210}{40*(\frac{1}{60})}-510](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B28210%7D%7B40%2A%28%5Cfrac%7B1%7D%7B60%7D%29%7D-510)
![w=120rpm](https://tex.z-dn.net/?f=w%3D120rpm)
Generally the equation for Time to come to rest is mathematically given by
![t=(\frac{\omega_0}{\omega_0-\omega})t](https://tex.z-dn.net/?f=t%3D%28%5Cfrac%7B%5Comega_0%7D%7B%5Comega_0-%5Comega%7D%29t)
![t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})](https://tex.z-dn.net/?f=t%3D%28%5Cfrac%7B510%7D%7B510-120rpm%7D%29%2840.0%29%28%5Cfrac%7B1%7D%7B60%7D%29)
![t=0.87min](https://tex.z-dn.net/?f=t%3D0.87min)
Therefore Angular displacement is
![\theta =(\frac{120+510}{2})0.87](https://tex.z-dn.net/?f=%5Ctheta%20%3D%28%5Cfrac%7B120%2B510%7D%7B2%7D%290.87)
![\theta=274rev](https://tex.z-dn.net/?f=%5Ctheta%3D274rev)